Transcript Electrochemistry

Chapter 18
Electrochemistry
Lesson 1
Electrochemistry
18.1 Balancing Oxidation–Reduction Reactions
18.2 Galvanic Cells
18.3 Standard Reduction Potentials
18.4 Cell Potential, Electrical Work, and Free
Energy
18.5 Dependence of Cell Potential on Concentration
18.6 Batteries
18.7 Corrosion
18.8 Electrolysis
18.9 Commercial Electrolytic Processes
Electrochemical Reactions
In electrochemical reactions, electrons are
transferred from one species to another.
Oxidation Numbers
In order to keep
track of what loses
electrons and what
gains them, we
assign oxidation
numbers.
Oxidation and Reduction
• A species is oxidized when it loses electrons.
– Here, zinc loses two electrons to go from neutral
zinc metal to the Zn2+ ion.
Oxidation and Reduction
• A species is reduced when it gains electrons.
– Here, each of the H+ gains an electron and they
combine to form H2.
Oxidation and Reduction
• What is reduced is the oxidizing agent.
– H+ oxidizes Zn by taking electrons from it.
• What is oxidized is the reducing agent.
– Zn reduces H+ by giving it electrons.
Assigning Oxidation Numbers
1. Elements in their elemental form have an
oxidation number of 0.
2. The oxidation number of a monatomic
ion is the same as its charge.
Assigning Oxidation Numbers
3. Nonmetals tend to have negative
oxidation numbers, although some are
positive in certain compounds or ions.
– Oxygen has an oxidation number of −2,
except in the peroxide ion in which it has
an oxidation number of −1.
– Hydrogen is −1 when bonded to a metal,
+1 when bonded to a nonmetal.
Assigning Oxidation Numbers
3. Nonmetals tend to have negative
oxidation numbers, although some are
positive in certain compounds or ions.
– Fluorine always has an oxidation number
of −1.
– The other halogens have an oxidation
number of −1 when they are negative;
they can have positive oxidation
numbers, however, most notably in
oxyanions.
Assigning Oxidation Numbers
4. The sum of the oxidation numbers in a
neutral compound is 0.
5. The sum of the oxidation numbers in a
polyatomic ion is the charge on the
ion.
Balancing Oxidation-Reduction
Equations
Perhaps the easiest way to balance the
equation of an oxidation-reduction
reaction is via the half-reaction method.
Balancing Oxidation-Reduction
Equations
This involves treating (on paper only) the
oxidation and reduction as two separate
processes, balancing these half reactions,
and then combining them to attain the
balanced equation for the overall reaction.
Half-Reaction Method
1. Assign oxidation numbers to determine
what is oxidized and what is reduced.
2. Write the oxidation and reduction halfreactions.
Half-Reaction Method
3. Balance each half-reaction.
a.
b.
c.
d.
Balance elements other than H and O.
Balance O by adding H2O.
Balance H by adding H+.
Balance charge by adding electrons.
4. Multiply the half-reactions by integers
so that the electrons gained and lost
are the same.
Half-Reaction Method
5. Add the half-reactions, subtracting
things that appear on both sides.
6. Make sure the equation is balanced
according to mass.
7. Make sure the equation is balanced
according to charge.
Half-Reaction Method
Consider the reaction between MnO4− and C2O42− :
MnO4−(aq) + C2O42−(aq)  Mn2+(aq) + CO2(aq)
Half-Reaction Method
First, we assign oxidation numbers.
+7
+3
+2
+4
MnO4− + C2O42-  Mn2+ + CO2
Since the manganese goes from +7 to +2, it is reduced.
Since the carbon goes from +3 to +4, it is oxidized.
Oxidation Half-Reaction
C2O42−  CO2
To balance the carbon, we add a
coefficient of 2:
C2O42−  2 CO2
Oxidation Half-Reaction
C2O42−  2 CO2
The oxygen is now balanced as well. To
balance the charge, we must add 2
electrons to the right side.
C2O42−  2 CO2 + 2 e−
Reduction Half-Reaction
MnO4−  Mn2+
The manganese is balanced; to balance
the oxygen, we must add 4 waters to the
right side.
MnO4−  Mn2+ + 4 H2O
Reduction Half-Reaction
MnO4−  Mn2+ + 4 H2O
To balance the hydrogen, we add 8 H+ to
the left side.
8 H+ + MnO4−  Mn2+ + 4 H2O
Reduction Half-Reaction
8 H+ + MnO4−  Mn2+ + 4 H2O
To balance the charge, we add 5 e− to the
left side.
5 e− + 8 H+ + MnO4−  Mn2+ + 4 H2O
Combining the Half-Reactions
Now we evaluate the two half-reactions
together:
C2O42−  2 CO2 + 2 e−
5 e− + 8 H+ + MnO4−  Mn2+ + 4 H2O
To attain the same number of electrons
on each side, we will multiply the first
reaction by 5 and the second by 2.
Combining the Half-Reactions
5 C2O42−  10 CO2 + 10 e−
10 e− + 16 H+ + 2 MnO4−  2 Mn2+ + 8 H2O
When we add these together, we get:
10 e− + 16 H+ + 2 MnO4− + 5 C2O42− 
2 Mn2+ + 8 H2O + 10 CO2 +10 e−
Combining the Half-Reactions
10 e− + 16 H+ + 2 MnO4− + 5 C2O42− 
2 Mn2+ + 8 H2O + 10 CO2 +10 e−
The only thing that appears on both sides are the
electrons. Subtracting them, we are left with:
16 H+ + 2 MnO4− + 5 C2O42− 
2 Mn2+ + 8 H2O + 10 CO2
The Half–Reaction Method for Balancing Redox
Equations in Basic Solution
1. Use the half–reaction method as specified for acidic
solutions to obtain the final balanced equation as if
H+ ions were present.
2. To both sides of the equation, add a number of OH–
ions that is equal to the number of H+ ions present.
(You want to eliminate H+ by turning is into H2O)
3. Form H2O on the side containing both H+ and OH–
ions, and eliminate the number of H2O molecules
that appear on both sides of the equation.
4. Check that elements and charges are balanced.
Balancing in Basic Solution
16 H+ + 2 MnO4− + 5 C2O42− 
2 Mn2+ + 8 H2O + 10 CO2
• How would our last answer look in basic
solution?
Galvanic Cell
• A device in which chemical energy is
converted to electrical energy.
• It uses a spontaneous redox reaction to
produce a current that can be used to generate
energy or to do work.
A Galvanic Cell
In Galvanic Cell:
• Oxidation occurs at the anode.
• Reduction occurs at the cathode.
• Salt bridge or porous disk allows ions to flow
without extensive mixing of the solutions.
 Salt bridge – contains a strong electrolyte held in a
gel–like matrix.
 Porous disk – contains tiny passages that allow
hindered flow of ions.
Galvanic Cell (Voltaic Cell)
Voltaic Cells
In spontaneous
oxidation-reduction
(redox) reactions,
electrons are
transferred and
energy is released.
Voltaic Cells
• We can use that
energy to do work if
we make the
electrons flow
through an external
device.
• We call such a
setup a voltaic cell
(Galvanic Cells).
Voltaic Cells
• A typical cell looks
like this.
• The oxidation
occurs at the anode.
• The reduction
occurs at the
cathode.
Voltaic Cells
Voltaic Cells
As one electron
flows from the
anode to the
cathode, the
charges in each
beaker would not be
balanced and the
flow of electrons
would stop.
Voltaic Cells
• Therefore, we use a
salt bridge, usually a
U-shaped tube that
contains a salt
solution, to keep the
charges balanced.
– Cations move toward
the cathode.
– Anions move toward
the anode.
Voltaic Cells
• In the cell, then,
electrons leave the
anode and flow
through the wire to
the cathode.
• As the electrons
leave the anode, the
cations formed
dissolve into the
solution in the
anode compartment.
Voltaic Cells
• As the electrons
reach the cathode,
cations in the
cathode are
attracted to the now
negative cathode.
• The electrons are
taken by the cation,
and the neutral
metal is deposited
on the cathode.
Electromotive Force (emf)
• Water only
spontaneously flows
one way in a
waterfall.
• Likewise, electrons
only spontaneously
flow one way in a
redox reaction—from
higher to lower
potential energy.
Electromotive Force (emf)
• The potential difference between the
anode and cathode in a cell is called the
electromotive force (emf).
• It is also called the cell potential, and is
designated Ecell.
Cell Potential
Cell potential is measured in volts (V).
J
1V=1
C
Standard Reduction Potentials
Reduction
potentials for
many
electrodes
have been
measured and
tabulated
(Table 18.1 on
p. 845).
Standard Hydrogen Electrode
• Their values are referenced to a standard
hydrogen electrode (SHE).
• By definition, the reduction potential for
hydrogen is 0 V:
2 H+ (aq, 1M) + 2 e−  H2 (g, 1 atm)
Standard Cell Potentials
The cell potential at standard conditions
can be found through this equation:
Ecell
 (cathode) − Ered
 (anode)
 = Ered
Because cell potential is based on
the potential energy per unit of
charge, it is an intensive property.
Cell Potentials
• For the oxidation in this cell,
Ered
 = −0.76 V
• For the reduction,
Ered
 = +0.34 V
Cell Potentials
Ecell
 = Ered
 (cathode) − Ered
 (anode)
= +0.34 V − (−0.76 V)
= +1.10 V
Oxidizing and Reducing Agents
• The strongest
oxidizers have the
most positive
reduction potentials.
• The strongest
reducers have the
most negative
reduction potentials.
Standard Reduction Potentials
• Values of E° in standard table of half-cell potentials
are for reduction half-cell reactions
 Table 18.1
 1 M, 1 atm, 25°C
• When a half-reaction is reversed, the algebraic sign of
E° changes.
• When a half-reaction is multiplied by an integer, the
value of E° remains the same.
• A galvanic cell runs spontaneously in the direction
that gives a positive value for E°cell
Example: Fe3+(aq) + Cu(s)  Cu2+(aq) + Fe2+(aq)
• Half-Reactions:
 Fe3+ + e–  Fe2+
 Cu2+ + 2e–  Cu
E° = 0.77 V
E° = 0.34 V
• To balance the cell reaction and calculate the
cell potential, we must reverse reaction 2.
 Cu  Cu2+ + 2e–
– E° = – 0.34 V
• Each Cu atom produces two electrons but each
Fe3+ ion accepts only one electron, therefore
reaction 1 must be multiplied by 2.
 2Fe3+ + 2e–  2Fe2+
E° = 0.77 V
Standard Cell Potential
2Fe3+ + 2e–  2Fe2+ ; E° = 0.77 V (cathode)
Cu  Cu2+ + 2e– ; – E° = – 0.34 V (anode)
• Balanced Cell Reaction:
Cu + 2Fe3+  Cu2+ + 2Fe2+
• Cell Potential: E
E°cell = E°(cathode) – E°(anode)
E°cell = 0.77 V – 0.34 V = 0.43 V
Cell Notations for Galvanic Cells
•
•
•
•
•
Used to describe electrochemical cells.
Anode components are listed on the left.
Cathode components are listed on the right.
Separated by double vertical lines.
The concentration of aqueous solutions should be
specified in the notation when known.
• Example: Mg(s)|Mg2+(aq)||Al3+(aq)|Al(s)
 Mg  Mg2+ + 2e– (anode)
 Al3+ + 3e–  Al
(cathode)
Description of a Galvanic Cell
• The cell potential is always positive for a galvanic
cell, where E°cell = E°(cathode half-cell) – E°(anode half-cell) (as
given in the table of half-cell potentials)
• Anode is the negative terminal and cathode is the
positive terminal.
• Electron flows from the anode (-) to cathode (+).
• Current flows from cathode(+) to anode(-).
• Positive ions flows from anode to cathode half-cells
and negative ions flows from cathode to anode
though the “salt bridge”.
Cell Potential, Free Energy, and Electrical Work
Maximum cell potential and free energy
• Directly related to the free energy difference between
the reactants and the products in the cell reaction.
 ΔG° = –nFE°
F = 96,485 C/mol e–
Work:
• In any real, spontaneous process some energy is
always lost (wasted) – the actual work obtained is
always less than the calculated (maximum) value.
Electrochemistry
There are two kinds electrochemical cells.
• Electrochemical cells containing
nonspontaneous chemical reactions are
called electrolytic cells.
• Electrochemical cells containing
spontaneous chemical reactions are
called voltaic or galvanic cells.
1. Galvanic cell (voltaic cell)— energy
released during a spontaneous
reaction (G < 0) generates electricity
2. Electrolytic cell—consumes
electrical energy from an external
source to cause a nonspontaneous
redox reaction to occur (G > 0)
3. The cathode is negative in
electrolytic cells and positive in
voltaic cells.
4. The anode is positive in electrolytic
cells and negative in voltaic cells.